Leader: Gun? What for?
Ed White: Heck, Leonov and Belyayev have landed, right? No need for weapons!
Leader: Very funny. So, anyway, the new suit has 18 layers, to protect against micrometeorites and extreme cold.
Assistant: As for the "gun," you wanted us to say it plain, so I don't call it a "handheld maneuvering unit." It fires air, not bullets.
Meanwhile, in Russia, Pavel Belyayev hears a chopping noise:
Pavel: Colonel Leonov! Hello!
Rescuer: OK, it's about 9 km to the helicopter. Are you up for some skiing? (The Colonel sighs.)
Alexei: Oy. We've just been around the world a few times... what's a few more kilometers? Let's just get out of here.
We now proceed with the Gemini IV launch. It is now 00:04:29:29 GET, which Ottaviani explains means 0 days, 4 hours, 29 minutes, 28 seconds since liftoff/
Mission Control: Gemini 4, KKKKKKXXXX Hawaii Capcom. What KXXX your status now? KXXX
Ed: About ready to start getting out. ...I'm separating from the spacecreaft.
(In Russia, the cosmonauts are greeted, "Why are you so overdressed, Blondie? You'd think it was cold out there or something!)
Mission Control: OK, you're right in front, Ed. You look beautiful.
Ed: I... I... I feel like a million dollars!
(Pffft! As Ottaviani explains, of course, the ZOT GUN didn't make any sound -- "pffft" or otherwise -- in the vacuum of space. Nothing does!)
Ed: The gun works real great, Jim.
Mission Control: Wait a second, let me take your picture.
(In Russia, a reporter asks, "Cosmonaut Leonov, what can you tell us about your flight and landing? Alexei responds, "Provided with a special suit, man can survive and work in outer space. That's all. Thank you for your attention.")
Mission Control: GeminKXXXXXX 4, Houston KXXX COM.
Jim: He's out, Gus, and it's really nifty. (Gus Grissom is in charge of Houston CAPCOM.)
Gus: Listen. Our vox doesn't work very well.
Ed: Hey Gus, do you read? KXXX
Gus: It's not really working, Ed.
Ed: No, it's very easy to maneuver with the gun. The only problem I have is that I've exhausted its fuel now. That is the greatest experience I've -- it's tremendous!
Gus (00:04:26:31 GET): You've got about KKKKXXX 5 minutes.
Jim: Hey, you smeared up my windshield, you dirty dog!
Ed: Did I really? Well, hand me a Kleenex and I'll clean it.
Jim: Ha.
Gus: GemiKXXX-XXXton CAPCOM. Gemini 4, KXXXX. (10 minutes later...) Gemini 4 Houston CAPCOM, Gemini 4 KXXX.
Jim: Oh yes, that's great, the clouds on water behind you.
Gus: Gemini 4, Houston!
Jim: Ah, let's see what the flight director has got to say. Gus, this is Jim. Got any messages for us?
Gus: Gemini 4, get back in!
Jim: Back in?
Gus: Roger, we've been trying to talk to you for a while.
Ed: Aw... let me just take a few pictures.
Gus: No, back in, come on.
Ed: I'm coming.
Gus (3 minutes later): Ed, Ed, come on in here! (2 minutes later) Come on... get back here before it gets dark.
Ed: This is the saddest moment of my life.
Gus: Are you getting him back in? OK. get him back in.
Jim: He's standing on the seat now. I can't read you. Say again?
Gus: You got you cabin lights KXXX bright in case KXXXXXX hit darkness? Is he getting back in?
Ed (finally in): That was something. That was the most natural feeling, Jim.
Now that Gemini 4 is complete, let's move on to the famous Apollo missions. As Ottaviani explains, NASA engineers and astronauts met their Soviet counterparts now and then. So they're sad when they read that Sergei Korolev had passed away:
American scientist: Yeah, he was some kinda engineer. All their guys talked about him, without telling us his name -- sometimes I wonder if even they knew it. You gotta wonder what we had up his sleeve. A sad day.
(T-minus 2 years, 5 months, 23 days, Apollo/Saturn 204 AS-204 mission checkout.)
Ed: How are ya?
Roger: Good to see you Ed. Hop on in and let's get this test going.
Ed: You bet! Watch your head there, Roger.
Roger (Chaffee = Apollo 1 Lunar Module Pilot): Thanks, Ed.
Ed: Age before beauty, Gus.
Gus (Grissom = Apollo 1 Commander): Ha. Funny.
Ed (White = Apollo 1 Command and Service Module Pilot): Okay... all in. Let's get this thing done -- I'm tired of these ground tests. I want to get back up there!
But Apollo 1 isn't a success but a disaster -- the crew members all die during a ground test. The next scene takes place three months later -- May 1st, 1967. T-minus 2 years, 2 months, 19 days, final meeting with NAA on the Apollo 1 accident.
Paup (on phone): ...Time of the accident we'd already made 39 of 45 changes the review board recommended, and the other 6 were in the works! And that first change notice, we told NASA about it at the Lunar Mode Decision Congerence.
Storms: I know, I know. But...
Paup: So... what are you saying, Stormy?
Storms: Pin the blame on NASA and everybody loses. North American has gotta keep building the CSM and Saturn's final stage. Can't meet the goal any other way. But... Congress and the American people are gonna demand changes, and they have a right to do it.
Joe Shea: Okay, everyone.
Storms: So...yeah. Only one choice.
Joe Shea: We're here to review... uh... to review the roles played by NASA and North American Aviation... and...
Paup: If I may, Joe. North American accepts its share of responsibility for the Apollo 1 incident, bt if we're going to make it there and back by the end of the decade, we all know we cannot start over. As a result, we think it is prudent...
Joe Shea: Enough.
Storms: Man. Moon. Decade.
We move forward a year and several Apollo missions. It is now 1968. Gene Kranz took over from Chris Kraft as Mission Control's "Flight."
Gene Kranz: Hey Chris, if Apollo 6 was A and Apollo 7 was B...
Chris Kraft: Okay, Gene, call it C-prime if you want. Whatever you do, when you talk to the press, it's a "High Earth Orbital" mission. But the reporters don't get more details until we have a flight plane so -- get goin'. I'll talk to Werner.
Werner (on phone in Huntsvile, Alabama): Yes, Hello, Chris. You want to do what? Let me think. No... never mind. Once you decide to put a man on top of the Saturn V... it doesn't matter how far you go. So you may as well go to the moon.
Gene (NASA -- Houston, Texas): Lunar orbit on Christmas? Sounds good to me. What's that gonna look like?
Chris: A 69-mile circular orbit.
Gene: Okay, then... Better watch out for that 70-mile high mountain when you come around the far side, then.
Chris: Very funny. But I don't think my wife will think so.
They are speaking with Frank Borman: Commander (CDR), Jim Lovell: Command Module Pilot (CMP), and Bill Anders: Lunar Module Pilot (LMP). This is a nice place to end our reading today -- it's my first post in December, so why not mention the Christmas lunar orbit mission?
Today on her Mathematics Calendar 2019, Theoni Pappas writes:
The shaded area of this rectangle is 20 - pi square units. If the rectangle's dimensions are integers, what is its width?
[Here is the given info from the diagram, the area not shaded is that of a circle whose diameter is the same as the width.]
There's a quick and dirty trick to this problem. As the area of the rectangle minus the circle is 20 - pi, this means that the rectangle's area must be 20 and the circle's area must be pi -- otherwise, the dimensions of the rectangle wouldn't be integers! Since the area of the circle is pi, it must be a circle of unit radius. Hence its diameter is 2 -- and that's the width of the rectangle. Therefore our desired width is 2 -- and of course, today's date is the second.
Of course this seems like cheating -- from A - B = C - D, we can't in general conclude A = C or B = D, so that rectangle - circle = 20 - pi does not necessarily imply rectangle = 20 or circle = pi. But we are given that the dimensions are integers, so rectangle = integer and circle = integer * pi. Thus I am justified in setting rectangle = 20 and circle = pi and proceeding from there.
In fact, it's possible to make this argument rigorous using Linear Algebra. We already know about the concept of a vector -- a vector space can be taken over any field. In high school Geometry, that field is assumed to be the field R of real numbers. But there's nothing stopping us from considering a vector space over another field, such as the field Q of rational numbers. And in fact, it happens that the set R of real numbers can be considered a vector space over Q.
Now in this vector space, the vectors 1 and pi are considered linearly independent over Q. This means that no rational multiple of 1 can ever equal pi, and vice versa. (The vectors <1, 0> and <0, 1> are likewise independent in our usual 2D vector space R^2.)
Why does linear independence matter? Well, we can prove that if x and y are linearly independent, then Ax + By = Cx + Dy does imply A = C and B = D. So we can let x and y be the vectors 1 and pi (in the vector space R), A be the area of the rectangle, and B the negative square of the radius:
Ax + By = 20 - pi
so that A = C = 20 and B = D = -1. From this -r^2 = -1 we conclude r = 1 and so the diameter and width are indeed 2. This is the only possible rational (much less integer) solution, since we are working in the field Q.
This argument also shows why in Algebra II, we can solve a complex equation by setting the real and imaginary parts equal to each other -- C is a 2D-vector space over R, and in this vector space, the vectors 1 and i are linearly independent.
On the other hand, some Pappas problems don't specify that the solution is an integer, so we'd be cheating if we just assume that it's an integer (because it's a date, and dates are integers). For example, instead of specifying that the dimensions are integers, Pappas could have told us that the length of the rectangle is 10. It would then be cheating to assume that rectangle = 20 and circle = pi in that situation -- instead we must write:
10(2r) - pi r^2 = 20 - pi
and solve this equation to obtain r = 1. (Then again, as a quadratic equation it has two solutions -- the other is r = 20/pi - 1, which is also valid. In this case, Pappas would add something like r < 2 or "the width is less than 4" to force r = 1 as the only solution.)
Today is Cyber Monday. I ordered my copy of the new Rebecca Rapoport math calendar today.
Meanwhile, today we return from Thanksgiving break by starting Chapter 7. Lesson 7-1 of the U of Chicago text is on drawing triangles as an introduction to triangle congruence.
This is what I wrote last year about today's lesson. Actually, I didn't say much about the lesson per se, but I did write extensively about proofs, the major theme for today's post:
I introduced the concept of low-, medium-, and high-level proofs. These categories aren't rigid, but here's an approximate division:
Low-level: Prove SAS Congruence from first principles (i.e. transformations, if it's Common Core)
Mid-level: Prove the Isosceles Triangle Theorem from SAS Congruence
High-level: Prove the Equilateral Triangle Theorem (i.e. that an equilateral triangle is equiangular) from the Isosceles Triangle Theorem
So we can somewhat see the difference among these levels -- in particular, we may use the lower-level theorems in the proofs of the higher-level theorems.
But there's a more important distinction among these levels in the Geometry classroom. Teachers are more likely to ask students to prove higher-level than lower-level theorems. Many Geometry texts, especially pre-Core, don't expect students to prove our low-level theorems, such as SAS Congruence from first principles. Indeed, they absolve students from the responsibility of proving SAS completely by making it a postulate!
And now we see where the opponents of Common Core come in. They point out that geometry based on transformations is too experimental to appear in the classroom. Instead, they favor the pre-Core status quo -- just declare SAS a postulate and throw out transformations altogether,
Now here's the problem with this thinking -- low-level is to mid-level as mid-level is to high-level. I can now imagine a hypothetical class where not only do we avoid the low-level derivation of SAS from transformations, but we can avoid the mid-level derivation of the Isosceles Triangle Theorem from SAS as well. Instead, just declare the Isosceles Triangle Property to be a postulate and throw out SAS Congruence altogether! Students can still prove interesting theorems from this Isosceles Triangle "Postulate," including the Equilateral Triangle Theorem -- even the first problem from that Weeks and Adkins page from two weeks ago can be proved using only the Isosceles Triangle "Postulate" (and its converse, which could be declared yet another postulate).
One may argue that no Geometry text actually does this -- but au contraire, there really is a text that does something like the above. In Lesson 5-1 (old version) of Michael Serra's Discovering Geometry, Conjecture 27 is the Isosceles Triangle Conjecture (and Conjecture 28 is its converse), while Conjecture 29 is the Equilateral Triangle Conjecture (stated as a biconditional), with a paragraph proof provided to show how 29 follows from 28. So even though all three of these statements are labeled as "conjectures," the net effect is that 27 and 28 are postulates (as no proof is given), while 29 is actually a theorem proved using postulate 28.
Of course, this may seem silly -- Serra doesn't avoid SAS altogether, but instead gives it later on in the same chapter (Lesson 5-4, old version). And ultimately when we reach the end of the book when two-column proofs are taught, students are asked to prove the Isosceles Triangle Theorem using SAS, just as in most other texts.
But it does show that the pre-Core status quo is attacked on two fronts. If you argue that students should learn SAS so that the Isosceles Triangle Property can be a theorem rather than a postulate, then why not take it further in the direction of more rigor, and teach the students about reflections so that SAS can be a theorem rather than a postulate? Or going the other way, if you argue that students shouldn't have to learn how SAS follows from reflections, then why not take it further and say that students shouldn't have to learn how the Isosceles Triangle Property follows from SAS? It's not at all obvious why the exact status quo (SAS a postulate, Isosceles Triangle Property a theorem) is neither excessively nor insufficiently rigorous.
Now there is an argument that, if true, would vindicate the status quo defenders. It could be that the mid-level derivation of the Isosceles Triangle Theorem from SAS is easy for the students to understand, thereby preparing them well for the rest of Geometry and subsequent courses, but the low-level derivation of SAS from reflections is too hard for them and makes them cry after the test, thereby discouraging them from taking subsequent math courses. If this can be demonstrated, then the status quo is exactly right. Then again, until such a demonstration can be made, it's just as likely that the Common Core transformation approach, or even Serra's conjecture approach, could be correct.
There's one more thing that we must take into consideration -- the distinction between the Common Core Standards and the Common Core tests. Much of all my chapter juggling has occurred because I'd originally set up my lessons to match the standards, only to see something else on the tests. The standards state that students should learn how SAS and the other congruence and similarity theorems follow from the properties of transformations -- but such low-level proofs don't appear on the PARCC or any other Common Core test. It's actually easier to test for medium- and high-level proofs on a test, and the PARCC is no exception.
The PARCC question where students have to derive the Alternate Interior Angles Consequence from the Corresponding Angles Consequence is a mid-level proof. The PARCC question where students have to prove that the sum of the exterior angles of a triangle is 360 is definitely a high-level proof (after all, the triangle sum ultimately goes back to alternate interior angles). A low-level proof in this tree would be to show how the Corresponding Angles Consequence goes back to transformations.
Mid-level proofs on the PARCC are problematic -- and it's these questions that drive most of the changes in my curriculum. We saw last week how although the Common Core Standards ask students to derive SAS Similarity from the properties of dilations, a PARCC question asks them to derive a mid-level property of dilations from SAS~ instead.
But high-level proofs cause the fewest curriculum problems. I consider the classic two-column proofs of U of Chicago's Lesson 7-3 -- where students use SAS to prove two triangles congruent, but the "S" comes from the Reflexive Property or the definition of midpoint and the "A" comes from the Vertical Angles Theorem or some other result -- to be high-level proofs. This is because they appear at the top of the proof tree, rather than branch out to be used in other theorems. Such proofs don't require the students to derive SAS as a theorem.
On this blog, I will present low-level proofs in worksheets in order to meet the Common Core Standards, but I don't expect students to reproduce them in the exercises or on a quiz or test. On the other hand, students will have to know and understand the mid- and high-level proofs.
Let's get to today's worksheet. Now as it turns out, not only did I begin Chapter 7 last year near the Thanksgiving break, but it was also when I was purchasing a new computer and working hard to get it installed and connected to the Internet. These two facts combined mean that I don't necessary have a great Lesson 7-1 worksheet from last year.
Last year I gave some sort of an activity, where students were given parts of a triangle such as SS, AA, SSS, AAA, and so on, to determine whether they are sufficient to determine a triangle. Then I would follow this with a discussion of the results followed by some "review" problems. Once again, juggling the lessons around means that the students would be "reviewing" concepts that I haven't taught this year yet, such as Triangle Inequality and Triangle Sum. (Notice that this are related to the SSS and AAA conditions, respectively.)
So I decided to keep the activity-like part of the lesson and replace the Triangle Inequality and Triangle Sum questions with some more information about isometries and polygon congruence. This is what the resulting worksheet looks like:
The shaded area of this rectangle is 20 - pi square units. If the rectangle's dimensions are integers, what is its width?
[Here is the given info from the diagram, the area not shaded is that of a circle whose diameter is the same as the width.]
There's a quick and dirty trick to this problem. As the area of the rectangle minus the circle is 20 - pi, this means that the rectangle's area must be 20 and the circle's area must be pi -- otherwise, the dimensions of the rectangle wouldn't be integers! Since the area of the circle is pi, it must be a circle of unit radius. Hence its diameter is 2 -- and that's the width of the rectangle. Therefore our desired width is 2 -- and of course, today's date is the second.
Of course this seems like cheating -- from A - B = C - D, we can't in general conclude A = C or B = D, so that rectangle - circle = 20 - pi does not necessarily imply rectangle = 20 or circle = pi. But we are given that the dimensions are integers, so rectangle = integer and circle = integer * pi. Thus I am justified in setting rectangle = 20 and circle = pi and proceeding from there.
In fact, it's possible to make this argument rigorous using Linear Algebra. We already know about the concept of a vector -- a vector space can be taken over any field. In high school Geometry, that field is assumed to be the field R of real numbers. But there's nothing stopping us from considering a vector space over another field, such as the field Q of rational numbers. And in fact, it happens that the set R of real numbers can be considered a vector space over Q.
Now in this vector space, the vectors 1 and pi are considered linearly independent over Q. This means that no rational multiple of 1 can ever equal pi, and vice versa. (The vectors <1, 0> and <0, 1> are likewise independent in our usual 2D vector space R^2.)
Why does linear independence matter? Well, we can prove that if x and y are linearly independent, then Ax + By = Cx + Dy does imply A = C and B = D. So we can let x and y be the vectors 1 and pi (in the vector space R), A be the area of the rectangle, and B the negative square of the radius:
Ax + By = 20 - pi
so that A = C = 20 and B = D = -1. From this -r^2 = -1 we conclude r = 1 and so the diameter and width are indeed 2. This is the only possible rational (much less integer) solution, since we are working in the field Q.
This argument also shows why in Algebra II, we can solve a complex equation by setting the real and imaginary parts equal to each other -- C is a 2D-vector space over R, and in this vector space, the vectors 1 and i are linearly independent.
On the other hand, some Pappas problems don't specify that the solution is an integer, so we'd be cheating if we just assume that it's an integer (because it's a date, and dates are integers). For example, instead of specifying that the dimensions are integers, Pappas could have told us that the length of the rectangle is 10. It would then be cheating to assume that rectangle = 20 and circle = pi in that situation -- instead we must write:
10(2r) - pi r^2 = 20 - pi
and solve this equation to obtain r = 1. (Then again, as a quadratic equation it has two solutions -- the other is r = 20/pi - 1, which is also valid. In this case, Pappas would add something like r < 2 or "the width is less than 4" to force r = 1 as the only solution.)
Today is Cyber Monday. I ordered my copy of the new Rebecca Rapoport math calendar today.
Meanwhile, today we return from Thanksgiving break by starting Chapter 7. Lesson 7-1 of the U of Chicago text is on drawing triangles as an introduction to triangle congruence.
This is what I wrote last year about today's lesson. Actually, I didn't say much about the lesson per se, but I did write extensively about proofs, the major theme for today's post:
I introduced the concept of low-, medium-, and high-level proofs. These categories aren't rigid, but here's an approximate division:
Low-level: Prove SAS Congruence from first principles (i.e. transformations, if it's Common Core)
Mid-level: Prove the Isosceles Triangle Theorem from SAS Congruence
High-level: Prove the Equilateral Triangle Theorem (i.e. that an equilateral triangle is equiangular) from the Isosceles Triangle Theorem
So we can somewhat see the difference among these levels -- in particular, we may use the lower-level theorems in the proofs of the higher-level theorems.
But there's a more important distinction among these levels in the Geometry classroom. Teachers are more likely to ask students to prove higher-level than lower-level theorems. Many Geometry texts, especially pre-Core, don't expect students to prove our low-level theorems, such as SAS Congruence from first principles. Indeed, they absolve students from the responsibility of proving SAS completely by making it a postulate!
And now we see where the opponents of Common Core come in. They point out that geometry based on transformations is too experimental to appear in the classroom. Instead, they favor the pre-Core status quo -- just declare SAS a postulate and throw out transformations altogether,
Now here's the problem with this thinking -- low-level is to mid-level as mid-level is to high-level. I can now imagine a hypothetical class where not only do we avoid the low-level derivation of SAS from transformations, but we can avoid the mid-level derivation of the Isosceles Triangle Theorem from SAS as well. Instead, just declare the Isosceles Triangle Property to be a postulate and throw out SAS Congruence altogether! Students can still prove interesting theorems from this Isosceles Triangle "Postulate," including the Equilateral Triangle Theorem -- even the first problem from that Weeks and Adkins page from two weeks ago can be proved using only the Isosceles Triangle "Postulate" (and its converse, which could be declared yet another postulate).
One may argue that no Geometry text actually does this -- but au contraire, there really is a text that does something like the above. In Lesson 5-1 (old version) of Michael Serra's Discovering Geometry, Conjecture 27 is the Isosceles Triangle Conjecture (and Conjecture 28 is its converse), while Conjecture 29 is the Equilateral Triangle Conjecture (stated as a biconditional), with a paragraph proof provided to show how 29 follows from 28. So even though all three of these statements are labeled as "conjectures," the net effect is that 27 and 28 are postulates (as no proof is given), while 29 is actually a theorem proved using postulate 28.
Of course, this may seem silly -- Serra doesn't avoid SAS altogether, but instead gives it later on in the same chapter (Lesson 5-4, old version). And ultimately when we reach the end of the book when two-column proofs are taught, students are asked to prove the Isosceles Triangle Theorem using SAS, just as in most other texts.
But it does show that the pre-Core status quo is attacked on two fronts. If you argue that students should learn SAS so that the Isosceles Triangle Property can be a theorem rather than a postulate, then why not take it further in the direction of more rigor, and teach the students about reflections so that SAS can be a theorem rather than a postulate? Or going the other way, if you argue that students shouldn't have to learn how SAS follows from reflections, then why not take it further and say that students shouldn't have to learn how the Isosceles Triangle Property follows from SAS? It's not at all obvious why the exact status quo (SAS a postulate, Isosceles Triangle Property a theorem) is neither excessively nor insufficiently rigorous.
Now there is an argument that, if true, would vindicate the status quo defenders. It could be that the mid-level derivation of the Isosceles Triangle Theorem from SAS is easy for the students to understand, thereby preparing them well for the rest of Geometry and subsequent courses, but the low-level derivation of SAS from reflections is too hard for them and makes them cry after the test, thereby discouraging them from taking subsequent math courses. If this can be demonstrated, then the status quo is exactly right. Then again, until such a demonstration can be made, it's just as likely that the Common Core transformation approach, or even Serra's conjecture approach, could be correct.
There's one more thing that we must take into consideration -- the distinction between the Common Core Standards and the Common Core tests. Much of all my chapter juggling has occurred because I'd originally set up my lessons to match the standards, only to see something else on the tests. The standards state that students should learn how SAS and the other congruence and similarity theorems follow from the properties of transformations -- but such low-level proofs don't appear on the PARCC or any other Common Core test. It's actually easier to test for medium- and high-level proofs on a test, and the PARCC is no exception.
The PARCC question where students have to derive the Alternate Interior Angles Consequence from the Corresponding Angles Consequence is a mid-level proof. The PARCC question where students have to prove that the sum of the exterior angles of a triangle is 360 is definitely a high-level proof (after all, the triangle sum ultimately goes back to alternate interior angles). A low-level proof in this tree would be to show how the Corresponding Angles Consequence goes back to transformations.
Mid-level proofs on the PARCC are problematic -- and it's these questions that drive most of the changes in my curriculum. We saw last week how although the Common Core Standards ask students to derive SAS Similarity from the properties of dilations, a PARCC question asks them to derive a mid-level property of dilations from SAS~ instead.
But high-level proofs cause the fewest curriculum problems. I consider the classic two-column proofs of U of Chicago's Lesson 7-3 -- where students use SAS to prove two triangles congruent, but the "S" comes from the Reflexive Property or the definition of midpoint and the "A" comes from the Vertical Angles Theorem or some other result -- to be high-level proofs. This is because they appear at the top of the proof tree, rather than branch out to be used in other theorems. Such proofs don't require the students to derive SAS as a theorem.
On this blog, I will present low-level proofs in worksheets in order to meet the Common Core Standards, but I don't expect students to reproduce them in the exercises or on a quiz or test. On the other hand, students will have to know and understand the mid- and high-level proofs.
Let's get to today's worksheet. Now as it turns out, not only did I begin Chapter 7 last year near the Thanksgiving break, but it was also when I was purchasing a new computer and working hard to get it installed and connected to the Internet. These two facts combined mean that I don't necessary have a great Lesson 7-1 worksheet from last year.
Last year I gave some sort of an activity, where students were given parts of a triangle such as SS, AA, SSS, AAA, and so on, to determine whether they are sufficient to determine a triangle. Then I would follow this with a discussion of the results followed by some "review" problems. Once again, juggling the lessons around means that the students would be "reviewing" concepts that I haven't taught this year yet, such as Triangle Inequality and Triangle Sum. (Notice that this are related to the SSS and AAA conditions, respectively.)
So I decided to keep the activity-like part of the lesson and replace the Triangle Inequality and Triangle Sum questions with some more information about isometries and polygon congruence. This is what the resulting worksheet looks like:
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